How can I show that trace is Invariant under the change of basis?

[alphaneutrino]

How can I show that trace is Invariant under the change of basis?

 

[CompuChip]

The easiest way is to explicitly express the change of basis as
A \to A' = B A B^{-1}
and use the cyclic property
\operatorname{Tr}(ABC) = \operatorname{Tr}(BCA) = \operatorname{Tr}(CAB)

Or you can use that
\operatorname{Tr}(A) = \sum \lambda
where \lambda are the eigenvalues, and show that these are basis invariant
(for example, show that if v is a corresponding eigenvector, that (B A B^{-1}) (B v B^{-1}) = \lambda v as well).

 

[alphaneutrino]

thanks CompuChip

 

[CompuChip]

Actually, upon reading back my answer, I noted a little error. Instead of
<br /> (B A B^{-1}) (B v B^{-1}) = \lambda v<br />
I should of course have said
<br /> (B A B^{-1}) (B v B^{-1}) = \lambda (B v B^{-1})<br />
which simply says that if A v = \lambda v[/tex] then A&amp;#039; v&amp;#039; = \lambda v&amp;#039; where the primes refer to matrices and vectors in the new basis (i.e. A&amp;#039; = B A B^{-1} and v&amp;#039; = B v B^{-1}).